keyycee
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what is the final concentration in molar when 0.3756mL of 4M NaCl is diluted to 100mL?

what I did was m = m/L
so I converted to ml values by diving by a 1000 then adding them then diving 4m by the value I got but its wrong
can you guys tell me step by step on how to do it
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Rather_Cynical
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I have no idea what you did, but one way you could do it is by first-principles (if you know units of concentration, you can go by its definitions and mathematical reasoning).

The units of concentration is usually expressed as moles per decimeter (moldm-3). The number of moles you've got in your initial amount = 0.0003756dm3 x 4moldm-3, which is 0.0015024mol.

Then you divide by the new volume of 0.1dm3 = 0.0015024mol/0.1dm-3 = 0.015024moldm-3.

The units for molar is also moldm-3, so your final answer would be 0.015024M or 0.0150M (3 significant figures)

EDIT - I'm assuming that you know 1mL could be written as 1cm3 or 0.001dm3, because volume of water is defined by exactly 1L = 1dm3
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keyycee
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(Original post by Rather_Cynical)
I have no idea what you did, but one way you could do it is by first-principles (if you know units of concentration, you can go by its definitions and mathematical reasoning).

The units of concentration is usually expressed as moles per decimeter (moldm-3). The number of moles you've got in your initial amount = 0.0003756dm3 x 4moldm-3, which is 0.0015024mol.

Then you divide by the new volume of 0.1dm3 = 0.0015024mol/0.1dm-3 = 0.015024moldm-3.

The units for molar is also moldm-3, so your final answer would be 0.015024M or 0.0150M (3 significant figures)

EDIT - I'm assuming that you know 1mL could be written as 1cm3 or 0.001dm3, because volume of water is defined by exactly 1L = 1dm3
thanks alot but which equation did u use?
also im stuck on another question
how many mmol are there in 10ml acetone?
MW=58gmol-1 0.791g/ML at 25 degrees celcius
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Rather_Cynical
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(Original post by keyycee)
thanks alot but which equation did u use?
also im stuck on another question
how many mmol are there in 10ml acetone?
MW=58gmol-1 0.791g/ML at 25 degrees celcius
I don't actually use the equations given at all, all I need to know is the units! If you're familiar with 1/2 x 2 = 1, then it should also logically make sense that 1mol/dm3 x 1dm3 = 1mol.

The concentration in molar multiplied by the volume in dm3 equals the number of moles, and the number of moles divided by new volume in dm3 equals concentration in molar.

The second question you've given me gives you a concentration and volume, so you should multiply those two to get an answer in grams. The number of moles would be the mass in grams divided by the molecular weight (no conversions necessary).

EDIT - misread concentration as density, I blame lack of sleep.
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keyycee
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(Original post by Rather_Cynical)
I don't actually use the equations given at all, all I need to know is the units! If you're familiar with 1/2 x 2 = 1, then it should also logically make sense that 1mol/dm3 x 1dm3 = 1mol.

The concentration in molar multiplied by the volume in dm3 equals the number of moles, and the number of moles divided by new volume in dm3 equals concentration in molar.

The second question you've given me gives you a density and volume, so you should multiply those two to get an answer in grams. The number of moles would be the mass in grams divided by the molecular weight (no conversions necessary).
the answer is wrong and the way ur saying is wrong too ur not using the right formula which is ideal gas equation
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Rather_Cynical
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If I recall, you asked for the final concentration after diluting a volume of solution of known concentration into a larger volume. The mathematical formulae is correct, it doesn't require the ideal gas equation (we're not trying to find pressure or gas constants here).

Neither is it required to find the number of moles using concentration and volume.
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keyycee
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(Original post by Rather_Cynical)
If I recall, you asked for the final concentration after diluting a volume of solution of known concentration into a larger volume. The mathematical formulae is correct, it doesn't require the ideal gas equation (we're not trying to find pressure or gas constants here).

Neither is it required to find the number of moles using concentration and volume.
but ive tried ur way but the answer is not correct its mutliple choice and i have the answrs it doesnt come out right
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Rather_Cynical
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That's odd - could you show me the question paper and mark scheme? I'd try to work it out, might not be understanding your question properly!
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keyycee
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(Original post by Rather_Cynical)
That's odd - could you show me the question paper and mark scheme? I'd try to work it out, might not be understanding your question properly!
there isnt a mark scheme i do pharmacy im in uni
the auestion is exactly how i have written but the answer is 136 mmol
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Rather_Cynical
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I think you might have been confused with what I meant, this is what I'm getting:

10ml x 0.791g per ml = 7.91g
7.91g divided by 58g per mol = 0.136mol
0.136mol x 1000 millimol per mol = 136 millimol
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fleur_de_haine
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(Original post by keyycee)
there isnt a mark scheme i do pharmacy im in uni
the auestion is exactly how i have written but the answer is 136 mmol
I think there are some wires being crossed here. If your first question asks for the final molar concentration the answer cannot be '136 mmol' as mol (and indeed mmol) is a unit of amount as opposed to concentration.

By my working the answer would be approximately 15mM. Can you show us how you approached the problem?

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Rather_Cynical
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(Original post by fleur_de_haine)
I think there are some wires being crossed here. If your first question asks for the final molar concentration the answer cannot be '136 mmol' as mol (and indeed mmol) is a unit of amount as opposed to concentration.

By my working the answer would be approximately 15mM. Can you show us how you approached the problem?

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Fleur, don't worry - it's a different question (s)he asked me if you follow the whole thread.
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fleur_de_haine
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(Original post by Rather_Cynical)
Fleur, don't worry - it's a different question (s)he asked me if you follow the whole thread.
Oh I see!

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